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A large pile of coins consists of pennies, nickels, dimes, and quarters (at least 30 of each). a. How many different collections of 30 coins can be chosen? b. What is the probability that a collection of 30 coins chosen at random will contain at least four coins of each type?

Short Answer

Expert verified
In summary, there are a total of 5456 different collections of 30 coins that can be chosen. The probability that a collection of 30 coins chosen at random will contain at least four coins of each type is approximately 12.46%.

Step by step solution

01

Part (a) - Finding the number of different collections of 30 coins

We can use the stars and bars method to determine the number of different collections of 30 coins. We have 4 types of coins, and we want to distribute 30 coins among these types. We have: - 3 bars to divide the coins into 4 types - 30 stars representing the coins Using the formula for stars and bars, which is given by \( C(n + k - 1, k - 1) \), where n is the number of stars (coins) and k is the number of bars, we get: \( C(30 + 3, 3) \) \( = C(33, 3) \) \( = \frac{33!}{3!(33-3)!} \) \( = \frac{33!}{3!30!} \) Calculate the value of the expression: \( = \frac{33 \times 32 \times 31}{3 \times 2 \times 1} \) \( = 5456 \) So, there are 5456 different collections of 30 coins.
02

Part (b) - Finding the probability that a collection of 30 coins chosen at random will contain at least four coins of each type

First, let's find the number of ways to distribute at least 4 coins of each type among the 30 coins. 1. If there are exactly 4 pennies, 4 nickels, 4 dimes, and 4 quarters, then we have 14 coins to distribute among the 4 types: \( C(14 + 3, 3) \) \( = C(17, 3) \) \( = \frac{17!}{3!14!} \) Calculate the value of the expression: \( = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} \) \( = 680 \) So, there are 680 ways to distribute 14 coins among the 4 types with at least four coins of each type. Now, in order to find the probability that a random collection of 30 coins will contain at least four coins of each type, we simply divide the number of ways to distribute the coins with at least four coins of each type by the total number of different collections of 30 coins: Probability = \(\frac{680}{5456}\) Calculate the probability: = 0.1246 (approximately) So, the probability that a collection of 30 coins chosen at random will contain at least four coins of each type is approximately 12.46%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stars and Bars Method
Understanding the stars and bars method in combinatorics can be a game changer when it comes to distributing objects into containers. This method is particularly helpful for solving problems where the order of selection does not matter, and we are dealing with identical objects being divided into distinct categories, such as coins of different denominations in a collection.

Imagine stars as items to be distributed (in our case, coins), and bars as dividers between different categories (types of coins). To determine the number of ways to distribute 30 identical coins into 4 distinct types (pennies, nickels, dimes, quarters), we arranged 30 stars with 3 bars interspersed to create 4 groups.

The mathematical formula embodying this visualization is given by \( C(n + k - 1, k - 1) \), where \( n \) represents the stars (coins) and \( k \) the bars (categories). For 30 coins and 4 types, this translates to \( C(33, 3) \) since we have 30 stars and 3 bars. This combinatorial calculation fetches us 5456, indicating the sheer variety of ways to combine 30 coins into 4 types.

Learning the stars and bars method equips students with a powerful tool for tackling a broad range of combinatorial problems beyond just coin collections, illuminating the paths across landscapes filled with discrete and seemingly indistinct choices.
Combinations and Permutations
The beauty of combinatorics lies in its ability to count without counting each possibility one by one. Combinations and permutations are two foundational concepts that facilitate this.

Combinations are used when the order does not matter, much like choosing a hand of cards from a deck. For our coin problem, we are dealing with combinations because we are interested in the groups of coins irrespective of the order they're in. The combinatorial formula \( C(n, k) = \frac{n!}{k!(n-k)!} \) where \( n \) is the total number of items to choose from and \( k \) is the number of items to choose, is the math behind how we solve these types of problems.

Permutations, on the other hand, consider order to be important, akin to arranging books on a shelf. However, in the context of our coin problem, permutations aren't pertinent since the sequence in which the coins are picked doesn't alter the outcome of interest.

Understanding when to use combinations or permutations is crucial. A small tip to remember is that combinations are silent on the sequence ('c' for choice, not concerning order), whereas permutations pronounce it ('p' for positions, prescribing sequence). This clarity eases the complexity when facing a variety of combinatorial challenges.
Probability
Probability is the measure of how likely an event is to occur, and it’s fundamental to making informed predictions based on mathematical principles. It's the cornerstone of games, weather forecasts, and even our day-to-day decision making.

In a scenario such as picking a random collection of 30 coins, probability aids in answering questions along the lines of 'What is the chance of drawing at least four of each type of coin?'. To calculate this, you would first enumerate the number of favorable outcomes (selecting at least four of each coin), then divide by the total number of possible outcomes (all different 30-coin combinations).

In our exercise, the probability was calculated by dividing the number of ways to distribute the coins with at least four of each type (680 combinations) by the total number of collections (5456 combinations), resulting in approximately 12.46%. This quantification of likelihood helps students grasp an abstract concept with concrete numbers, empowering them to predict outcomes and make decisions based on statistical significance rather than guesswork.

Whether it's about taking a chance in a board game or making predictions in more sophisticated fields like finance or meteorology, understanding probability is an invaluable asset.

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Most popular questions from this chapter

An instructor gives an exam with twelve questions. Students are allowed to choose any ten to answer. a. How many different choices of ten questions are there? b. Suppose five questions require proof and seven do not. (i) How many groups of ten questions contain four that require proof and six that do not? (ii) How many groups of ten questions contain at least one that requires proof? (iii) How many groups of ten questions contain at most three that require proof? c. Suppose the exam instructions specify that at most one of questions 1 and 2 may be included among the ten. How many different choices of ten questions are there? d. Suppose the exam instructions specify that either both questions 1 and 2 are to be included among the ten or neither is to be included. How many different choices of ten questions are there?

Suppose there are three routes from North Point to Boulder Creek, two routes from Boulder Creek to Beaver Dam, two routes from Beaver Dam to Star Lake, and four routes directly from Boulder Creek to Star Lake. (Draw a sketch.) a. How many routes from North Point to Star Lake pass through Beaver Dam? b. How many routes from North Point to Star Lake bypass Beaver Dam?

Suppose that a coin is tossed three times and the side showing face up on each toss is noted. Suppose also that on each toss heads and tails are equally likely. Let \(H H T\) indicate the outcome heads on the first two tosses and tails on the third, THT the outcome tails on the first and third tosses and heads on the second, and so forth. a. List the eight elements in the sample space whose outcomes are all the possible head-tail sequences obtained in the three tosses. b. Write each of the following events as a set and find its probability: (i) The event that exactly one toss results in a head. (ii) The event that at least two tosses result in a head. (iii) The event that no head is obtained.

A coin is loaded so that the probability of heads is \(0.7\) and the probability of tails is \(0.3\). Suppose that the coin is tossed twice and that the results of the tosses are independent. a. What is the probability of obtaining exactly two heads? b. What is the probability of obtaining exactly one head? c. What is the probability of obtaining no heads? d. What is the probability of obtaining at least one head?

A camera shop stocks eight different types of batteries. a. How many ways can a total inventory of 30 batteries be distributed among the eight different types? b. Assuming that one of the types of batteries is A76, how many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four A76 batteries? c. If an inventory of 30 batteries is selected at random from the cight different types, what is the probability that at least four A76 batteries will be included? d. If an inventory of 30 batteries is selected at random from the eight different types, what is the probability that exactly four A 76 batteries will be included?

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